Question

A certain weak base has a Kb of 7.40 × 10-7. What concentration of this base will produce a pH of 10.07?

Answer 1

Answer – Given, Ph = 10.07 , Kb = 7.40*10^-7

We know,

pH + pOH = 14

so, pOH = 14-pH

              = 14 -10.07

              = 3.93

So [OH^-] = 10^-pOH

                = 10^-3.93

                = 1.17*10^-4 M

So at equilibrium [OH^-] = x = 1.17*10^-4 M

We assume weak base is B

      B + H₂O <------> BH⁺ + OH^-

I    Y                         0       0

C   -x                      +x     +x

E Y-x                    +x    1.17*10^-4 M

So, Kb = [BH⁺] [ OH^-] / [B]

7.40*10^-7 = x *x /(Y-x)

7.40*10^-7 = 1.17*10^-4 M * 1.17*10^-4 M / ( Y -1.17*10^-4 M)

Since Kb is too small then we can neglect x in the (Y-x)

So, 7.40*10^-7 = 1.17*10^-4 M * 1.17*10^-4 M / Y

So, Y = (1.17*10^-4 M * 1.17*10^-4) / 7.40*10^-7

          = 0.0186 M

So [B] = 0.0186 M

So, 0.0186 M concentration of this base will produce a pH of 10.07